Thu gọn các tổng sau :
A=2 + 22 + 23 + 24 + ...+ 2100
B= 4 + 42 + 43 + .... + 4n
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a: Ta có: \(A=2+2^2+2^3+2^4+...+2^{100}\)
\(=2\left(1+2\right)+2^3\left(1+2\right)+...+2^{99}\left(1+2\right)\)
\(=3\cdot\left(2+2^3+...+2^{99}\right)⋮3\)
b: Ta có: \(B=4+4^2+4^3+...+4^{2022}\)
\(=4\left(1+4\right)+4^3\left(1+4\right)+...+4^{2021}\left(1+4\right)\)
\(=5\cdot\left(4+4^3+...+4^{2021}\right)⋮5\)
\(A=2+2^2+2^3+2^4+...+2^{99}+2^{100}\)
\(\Rightarrow2A=2^2+2^3+2^4+...+2^{100}+2^{101}\)
\(\Rightarrow A=2A-A=2^2+2^3+2^4+...+2^{100}+2^{101}-2-2^2-2^3-2^4-...-2^{99}-2^{100}=2^{101}-2\)
Ta có:
A = 2 + 22 + 23 + … + 22017
2A = 2.( 2 + 22 + 23 + … + 22017)
2A = 22 + 23 + 24 + … + 22018
2A – A = (22 + 23 + 24 + … + 22018) – (2 + 22 + 23 + … + 22017)
Vậy A = 22018 – 2
Ta có: A = 2 + 22 + 23 + … + 22017
2A = 2.( 2 + 22 + 23 + … + 22017)
2A = 22 + 23 + 24 + … + 22018
2A – A = (22 + 23 + 24 + … + 22018) – (2 + 22 + 23 + … + 22017)
A = 22018 – 2
Vậy A = 22018 – 2
\(a,2^2=4,2^3=8,2^4=16,2^5=32,2^6=64,2^7=128,2^8=256,2^9=512,2^{10}=1024\)
\(b,3^2=9,3^3=27,3^4=81,3^5=243\)
\(c,4^2=16,4^3=64,4^4=256\)
\(d,5^2=25,5^3=125,5^4=625\)
a) \(A=2+2^2+2^3+...+2^{2017}\)
\(2A=2^2+2^3+2^4+...+2^{2018}\)
\(2A-A=\left(2^2+2^3+2^4+...+2^{2018}\right)-\left(2+2^2+2^3+...+2^{2017}\right)\)
\(A=2^{2018}-2\)
b) \(C=1+3^2+3^4+...+3^{2018}\)
\(3^2\cdot C=3^2+3^4+3^6+...+3^{2020}\)
\(9C-C=\left(3^2+3^4+3^6+...+3^{2020}\right)-\left(1+3^2+3^4+...+3^{2018}\right)\)
\(8C=3^{2020}-1\)
\(\Rightarrow C=\dfrac{3^{2020}-1}{8}\)
\(Toru\)
a) 4 ; 8 ; 16 ; 32 ; 64
b) 9 ; 27 ; 81 ; 243
c) 16 ; 64 ; 256
d) 25 ; 125
Chúc bạn học tốt!! ^^
a) \(2^2=4\)
\(2^3=8\)
\(2^4=16\)
\(2^5=32\)
\(2^6=64\)
b) \(3^2=3\)
\(3^3=27\)
\(3^4=81\)
\(3^5=243\)
c) \(4^2=16\)
\(4^3=64\)
\(4^4=256\)
d) \(5^2=25\)
\(5^3=125\)
a.
$S=1+2+2^2+2^3+...+2^{2017}$
$2S=2+2^2+2^3+2^4+...+2^{2018}$
$\Rightarrow 2S-S=(2+2^2+2^3+2^4+...+2^{2018}) - (1+2+2^2+2^3+...+2^{2017})$
$\Rightarrow S=2^{2018}-1$
b.
$S=3+3^2+3^3+...+3^{2017}$
$3S=3^2+3^3+3^4+...+3^{2018}$
$\Rightarrow 3S-S=(3^2+3^3+3^4+...+3^{2018})-(3+3^2+3^3+...+3^{2017})$
$\Rightarrow 2S=3^{2018}-3$
$\Rightarrow S=\frac{3^{2018}-3}{2}$
Câu c, d bạn làm tương tự a,b.
c. Nhân S với 4. Kết quả: $S=\frac{4^{2018}-4}{3}$
d. Nhân S với 5. Kết quả: $S=\frac{5^{2018}-5}{4}$
a) \(S=1+2+2^2+2^3+...+2^{2022}=\dfrac{2^{2022+1}-1}{2-1}=2^{2023}-1\)
b) \(S=1+4+4^2+4^3+...+4^{2022}=\dfrac{4^{2022+1}-1}{4-1}=\dfrac{4^{2023}-1}{3}\)
\(S=1+2+2^2+2^3+...+2^{2022}\\ 2S=2+2^2+2^3+2^4+...+2^{2023}\\ 2S-S=2+2^2+2^3+2^4+...+2^{2023}-1-2-2^2-2^3-...-2^{2022}\\ S=2^{2023}-1\\ S=4+4^2+4^3+...+4^{2022}\\ 4S=4^2+4^3+4^4+...+4^{2023}\\ 4S-S=4^2+4^3+4^4+...+4^{2023}-4-4^2-4^3-...-4^{2023}\\ 3S=4^{2023}-4\\ S=\dfrac{4^{2023}-4}{3}\)
D=4+4^2+...+4^n
=>\(4\cdot D=4^2+4^3+...+4^{n+1}\)
=>\(3D=4^{n+1}+4^n+...+4^3+4^2-4^n-...-4^2-4\)
=>\(3D=4^{n+1}-4\)
=>\(D=\dfrac{4^{n+1}-4}{3}\)